What is the integral of (x-4)/(x^2+4x+3) ?

The integral of (x-4)/(x^2+4x+3) is 1/2 ln|x^2+4x+3|+3ln|x+3|-3ln|x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-4)/(x^2+4x+3)

\int \frac{x - 4}{x ^{2} + 4 x + 3} d x

\frac{1}{2} ln x^{2} + 4 x + 3 + 3 ln ∣ x + 3 ∣ - 3 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{x - 4}{x ^{2} + 4 x + 3} d x

Expand

\frac{x - 4}{x ^{2} + 4 x + 3} : \frac{x}{x ^{2} + 4 x + 3} - \frac{4}{x ^{2} + 4 x + 3}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ^{2} + 4 x + 3} d x - \int \frac{4}{x ^{2} + 4 x + 3} d x

\int \frac{x}{x ^{2} + 4 x + 3} d x = \frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣

\int \frac{4}{x ^{2} + 4 x + 3} d x = 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣)

= \frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ - 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣)

Simplify

\frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ - 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣) : \frac{1}{2} ln x^{2} + 4 x + 3 + 3 ln ∣ x + 3 ∣ - 3 ln ∣ x + 1 ∣

= \frac{1}{2} ln x^{2} + 4 x + 3 + 3 ln ∣ x + 3 ∣ - 3 ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 3 + 3 ln ∣ x + 3 ∣ - 3 ln ∣ x + 1 ∣ + C

\int 4 e^{- s t} d t

\int 5 sin (x^{4}) d x

\int - \frac{30}{x ^{2}} d x

\int (5 x^{2} + 6 x^{- 2}) d x

\int \frac{1}{x ln ^{p} ( x )} d x

What is the integral of (x-4)/(x^2+4x+3) ?
The integral of (x-4)/(x^2+4x+3) is 1/2 ln|x^2+4x+3|+3ln|x+3|-3ln|x+1|+C